What is the equilibrium constant of pure water at 25°C?

The equilibrium reaction, causes the water dissociation constant, Kw, is 1.01 × 10-14 at 25 °C. That is because every H+ (H3O+) ion these forms accompanied by the formation of an OH- ion, are the concentrations of these ions and in pure water the same thing can be calculated from Kw.

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IlaMends IlaMends · Answer 2 · 2019-04-02T07:20:27+02:00

Answer:The equilibrium constant of pure water is [tex]K_{eq}=1\times 10^{-14}[/tex]

Explanation:

[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]

[tex]K_{eq}=\frac{[H_3O^+][OH^-]}{[H_2O][H_2O]}[/tex]

[tex]K_i=K_{eq}\times [H_2O]}=\frac{[H_3O^+][OH^-]}{[H_2O]}[/tex]

[tex]K_i=\text{Ionization constant of water}[/tex]

Since,the water is found to be poorly ionized ,concentration of pure water practically remains the same. So, concentration of water can be combined with [tex]K_i[/tex] to give new constant known as ionic product of water that is [tex]K_w[/tex].

[tex]K_w=K_i\times [H_2O]=[H_3O^+][OH^-][/tex]

In pure water:

[tex][H_3O^+]=1\times 10^{-7} M]=[OH^-][/tex]

[tex]K_w=1\times 10^{-14}M^2[/tex]

[tex]K_w=K_{eq}=1\times 10^{-14}[/tex]